Name: SOLUTIONS

Math 241 F1H
Midterm Exam 3 Solutions
Prof. A.J. Hildebrand
Name: SOLUTIONS
Instructions:
1. No calculators, books, notes, formula sheets, etc.
2. Read the problems carefully. All the information you need to solve a problem is given in
the statement of the problem. By the same token, you should work the problem using only the
assumptions given in the problem; you cannot add assumptions on your own.
3. Show all work. An answer alone, without justification, will not earn credit. Write up your
solution in a logical and mathematically meaningful manner, using correct notation. In particular,
use arrow notation to denote vectors, and indicate dot products by a clearly visible dot (e.g., •). For
computational problems, box your final answer. You can leave numerical answers in “raw” form,
√
√
such as 3/11, or 3/2, or cos(1/ 241). However, you should simplify as much as possible, and, in
particular, evaluate standard trig values such as sin(π/4).
4. Cross out anything you don’t want to be considered for grading. For example, if you
started out on the wrong track, but caught your error and found the correct approach, cross out all
bad work.
5. Time: 50 minutes, plus possibly a small amount of grace period. This works out to about 10
minutes per problem. Budget your time wisely. Several problems just ask for the statement of
a formula/theorem; these should take very little time provided you know the formula or result in
question, and you might want to do these first. Of the computational problems, none requires
excessive calculations; if you get entangled in a messy computation, you are likely on the wrong
track, and you should move on to another problem. When you are finished, double-check your work.
Make sure you answered all questions, boxed your answers, provided justifications/explanations if
needed, and crossed out any bad work.
GOOD LUCK!
Problem (Pts)
Points
1 (10)
2 (15)
3 (15)
4 (20)
5 (20)
6 (20)
Total (100)
Math 241 F1H
Midterm Exam 3 Solutions
Prof. A.J. Hildebrand
R∞ 1 2
1. Evaluate the integral I = 0 e− 2 x dx by considering its square, I · I, and evaluating the latter as a
double integral. (Note that the range of integration is restricted to positive values of x.)
R∞
2
Solution: This is a variation on the Gaussian integral −∞ e−x dx, which was evaluated
in class as an illustration of double integrals in polar coordinates. The trick is to start out
with I · I, convert this integral to a double integral over the region 0 ≤ x < ∞, 0 ≤ y < ∞
i.e., the first quadrant, then evaluate this double integral by changing to polar coordinates.
Z ∞
Z ∞
2
y2
− x2
e− 2 dy
dx
e
I ·I =
y=0
x=0
∞ Z ∞
Z
e
=
Z
π/2
Z
∞
=
=
Z
1
2
e− 2 r rdrdθ
(convert to polar coordinates)
r=0
θ=0
π
2
dxdy
x=0
y=0
=
− 12 (x2 +y 2 )
∞
e−u du
(set u = r2 /2, du = rdr)
0
r
π −u ∞
π
−e
= ,
u=0
2
2
I=
π
2
Remark. This integral cannot be done by the usual tricks, e.g., integration by parts,
etc. The only way to do it (with the tools available at the calculus level) is to convert
it to a double integral as above, and then evaluate that integral by conversion to polar
coordinates. In obtaining the correct double integral, it is crucial to use y as the second
2
2
2
2
2
variable; otherwise the integrand would be e−x /2 · e−x /2 = e−x instead of e−(x +y )/2 ,
and the conversion to polar coordinates would not work.
1
Math 241 F1H
Midterm Exam 3 Solutions
Prof. A.J. Hildebrand
2. Suppose X and Y are random variables with joint density function
(
Ce−x−4y if x ≥ 0 and y ≥ 0,
f (x, y) =
0
otherwise,
where C is a constant.
(a) Determine the constant C.
Solution: For f (x, y) to be a probability density, the integral over the full range must
be equal to 1. Thus:
Z ∞ Z ∞
Ce−x−4y dydx
1=
x=0
y=0
∞
1 −4y
= C −e−x ]∞
−
e
x=0
4
y=0
=
C
.
4
C=4
(b) Find the probability P (X + Y ≤ 1).
Solution: The probability P (X + Y ≤ 1) is given by the double integral of f (x, y)
over the region in the first quadrant where x + y ≤ 1:
Z 1 Z 1−x
P (X + Y ≤ 1) =
4e−x−4y dydx
x=0
Z
y=0
1
=
e
−x
x=0
Z
−1 −4y
e
4
4
y=1−x
dx
y=0
1
=−
e3x−4 − e−x dx
x=0
1
= − e3x−4 + e−x
3
1
x=0
1
4
= − e−1 + 1 + e−4
3
3
2
Math 241 F1H
Midterm Exam 3 Solutions
Prof. A.J. Hildebrand
3. Consider the transformation of variables in R3 defined by x = u2 , y = v 3 , z = w4 .
(a) Compute the Jacobian determinant
∂(x, y, z)
of the above transformation.
∂(u, v, w)
Solution: The Jacobian of this transformation is
2u 0
0 ∂(x, y, z)
0 = 24uv 2 w3 .
= 0 3v 2
∂(u, v, w) 0
0 4w3 (b) Using the above transformation, set up, but do not evaluate, a triple integral in the variables
u, v, w, that gives the volume of the region (in the first octant) that is bounded by the surface
1/2
1/3
x
+ z 1/4 = 1 and the coordinate planes. Your answer should be of the form V =
R ∗ R ∗+R y∗
∗ dwdvdu, with appropriate expressions in place of the 7 asterisks.
∗ ∗ ∗
Solution: Setting u = x1/2 , v = y 1/3 , w = z 1/4 , the given surface becomes u+w +w =
1, and the region bounded by this surface and the coordinate planes (i.e., the planes
x = 0, y = 0, z = 0, or equivalently u = 0, v = 0, w = 0) is given by the inequalitites
0 ≤ u ≤ 1, 0 ≤ v ≤ 1 − u, 0 ≤ w ≤ 1 − u − w. Keeping in mind the extra factor
from the Jacobian determinant computed in the first part, 8uvw, the integral giving
the volume is
Z 1 Z 1−u Z 1−u−v
24uv 2 w3 dwdvdu
V =
u=0
v=0
w=0
3
Math 241 F1H
Midterm Exam 3 Solutions
Prof. A.J. Hildebrand
4. Quickies. The following problems are independent of each other.
(a) Convert the equation ρ = cos φ into an equation in rectangular coordinates, simplifying as
much as possible. (In particular, the equation should not involve inverse trig functions.)
Solution: Multiply by ρ to get ρ2 = ρ cos φ, which converts to x2 + y 2 + z 2 = z . or
x2 + y 2 + (z − 1/2)2 = 1/4 , which is the is the equation of a sphere with radius 1/2
and center (0, 0, 1/2).
R
(b) Express the vector line integral C F~ · d~r as a line integral with respect to arclength ds, i.e., in
R∗
the form ∗ ∗ ds, with appropriate expressions in place of the asterisks. Be sure to use correct
notation. (As usual, C denotes a smooth curve parametrized by ~r(t), a ≤ t ≤ b, and F~ = hP, Qi
a vector field in R2 whose component functions P and Q have continuous partial derivatives.)
Z
Z
F~ · T~ ds .
F~ · d~r =
Solution: The formula is listed on the Line Integral Handout:
C
C
(c) Find the volume of the image of the n-dimensional unit cube under the transformation ~x = T (~u)
from Rn to Rn defined by
x1 = 2u1 ,
x2 = 2u2 − u1
x3 = 2u3 − u2 ,
...
xn = 2un − un−1 ,
Be sure to show and explain your work.
Solution: By the general theory of linear transformations (see the handout “Linear
Algebra II”), the volume of the image of the unit cube is the “blow-up factor” for
volumes and equal to the absolute value of the determinant
of the associated matrix.
In the given case, this matrix is a triangular matrix
det A = 2 · 2 · · · 2 = 2n . (No integrations needed!)
4
2
−1
...
0
0
2
...
...
0 0
0 0
...
0 −1
...
...
...
2
, with determinant
Math 241 F1H
Midterm Exam 3 Solutions
Prof. A.J. Hildebrand
5. Let a, b, c, d be real numbers, and let F~ be the vector field in R2 defined by
F~ (x, y) = hax + by, cx + dyi.
(a) Find the work performed under the field F~ when moving an object along a straight line path
from (0, 0) to (2, 3).
Solution: The line segment can be parametrized by
~r 0 (t) = h2, 3i,
~r(t) = h2t, 3ti,
0 ≤ t ≤ 1.
With this parametrization we have
Z
Z 1
~
F · d~r =
((a(2t) + b(3t))(2) + (c(2t) + d(3t))(3))dt
C
0
Z
=
0
1
9
(4a + 6b + 6c + 9d)tdt = 2a + 3b + 3c + d
2
Remark: Green’s theorem cannot be applied here since the path is not a closed loop.
(b) Under what conditions on the values of a, b, c, d does the field F~ have a potential? Justify your
answer.
Solution:
Since F~ is defined with continuous partial derivatives in all of R2 , a
potential exists, i.e., the field is conservative, if and only if F~ satisfies the mixed partials
test, Py = Qx : Py = (ax + by)y = b, Qx = (cx + dy)x = c, so Py = Qx if only b = c
(c) Using a systematic approach (not by guessing or by trial and error), find a potential for F~ in
those cases in which a potential exists (i.e., the cases that meet the conditions of the previous
part).
Solution: Assume the condition from part (b), holds, i.e., b = c. Then we seek
f (x, y) such that (1) fx = ax + by and (2) fy = cx + dy = bx + dy. From (1),
f (x, y) = (1/2)ax2 +bxy+g(y). Then fy = bx+g 0 (y), and from (2), bx+g 0 (y) = bx+dy,
so g(y) = (1/2)dy 2 + C, where C is a constant. Substituting this into the formula for
f (x, y) gives f (x, y) = (1/2)ax2 + bxy + (1/2)dy 2 + C as a potential for F~ .
5
Math 241 F1H
Midterm Exam 3 Solutions
Prof. A.J. Hildebrand
6. Setup of integrals. Let E denote the solid region in R3 defined by
−2 ≤ x ≤ 2,
p
0 ≤ y ≤ 4 − x2 ,
p
p
x2 + y 2 ≤ z ≤ 8 − x2 − y 2 .
For each of the following problems, set up, but do not evaluate, an iterated integral of the requested
type in the coordinate system specified Your answer should be in the form
Z∗ Z∗ Z∗
Z∗ Z∗
?d ∗ d ∗
∗
? d ∗ d ∗ d∗
or
∗
∗
∗
∗
with appropriate expressions in place of the asterisks. (Hint: You may want to sketch the appropriate
regions to determine the correct integration limits.)
(a) The volume of E as a double integral in rectangular coordinates.
Solution:
Z
2
x=−2
√
Z
4−x2
p
p
( 8 − x2 − y 2 − x2 + y 2 )dydx.
y=0
(b) The volume of E as a triple integral in cylindrical coordinates.
Solution: The xy-limits describe a semi-disk, namely the part of the disk of radius
2 centered at the origin that lies above the x-axis). In polar coordinates, this disk is
given√by 0 ≤ θ ≤ π, 0 ≤ r ≤ 2. Using x2 + y 2 = r2 , the z-limits are from z = r to
z = 8 − r2 . Thus, in cylindrical coordinates, the integral becomes
Z
π
Z
2
Z
√
8−r 2
r dz dr dθ.
θ=0
r=0
z=r
(c) The volume of E as a triple integral in spherical coordinates.
Solution: To get limits in spherical coordinates,
we sketch the zr-region described by
√
off bounds on the spherical
the inequalities 0 ≤ r ≤ 2 and r ≤ z ≤ 8 − r2 . Reading
√
coordinates ρ and φ (see picture): φ ≤ π/4 and ρ ≤ 8. Thus, in spherical coordinates,
the integral becomes
Z π Z π/4 Z √8
ρ2 sin φ dρ dφ dθ.
θ=0
φ=0
ρ=0
6

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